Question

If $\cos \alpha =\frac{3\cos \beta -1}{3-\cos \beta }$ where $0<\beta <\alpha <\pi$, then the value of $\frac{\tan \frac{\alpha }{2}}{tan\frac{\beta }{2}}=$

• A. $1$
• B. $2$
• C. $\sqrt{3}$
• D. $\sqrt{2}$

Answer 1

Answer: (D)

$\sqrt{2}$

$\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}-\frac{3\left(\frac{1-{\tan }^2\frac{\beta }{2}}{1+{\tan }^2\frac{\beta }{2}}\right)-1}{3-\left(\frac{1-{\tan }^2\frac{\beta }{2}}{1+{\tan }^2\frac{\beta }{2}}\right)}=\frac{1-2{\tan }^2\frac{\beta }{2}}{1+2{\tan }^2\frac{\beta }{2}}$

$\therefore {\tan }^2\frac{\alpha }{2}=2{\tan }^2\frac{\beta }{2}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{2}\tan \frac{\beta }{2}\therefore 0<\beta <\alpha <\pi$